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Re: capacitors



I wrote:

Joe Scherrer wrote:
Now to my problem. If the capacitor remains connected to the voltage
source, the slab must now be pushed in because U(final)>U(initial).

Isn't the energy constant? When the conducting slab is in, the volume
with significant E-field is reduced, but where it remains it is
stronger. These two effects exactly cancel out, so that U ~ Int(E^2) is
constant.

Oops scratch that. Int(|E|) would be constant, but not Int(E^2). Doing
physics while short on sleep may be hazardous, but it beats operating
heavy equipment.

--
--James McLean
jmclean@chem.ucsd.edu
post doc, UC San Diego, Chemistry
moving to SUNY Geneseo Physics this fall